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Ex algebra  (7)
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Ex algebra (7)

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  • 1. Lista 4 de C´alculo I 2010-2 7 UFES - Universidade Federal do Esp´ırito Santo DMAT - Departamento de Matem´atica LISTA 4 - 2010-2 Limite infinito e no infinito Teoremas do confronto e anulamento Limites trigonom´etricos Nos exerc´ıcios 1. a 4. os gr´aficos de g e h s˜ao dados. Ache os limites laterais de f no ponto indicado. 1. f(x) = g(x) h(x) , no ponto x = 2 h g y x –2 0 2 4 –2 2 4 6 8 2. f(x) = g(x) h(x) , no ponto x = 3 g y x –2 2 4 –3 –2 –1 1 2 3 4 5 6 7 8 h y x –100 0 100 200 300 –2 2 4 6 8 3. f(x) = g(x) h(x) , no ponto x = 2 g y x –1 1 2 3 4 5 6 –2 2 4 h y x –300 –200 –100 0 100 200 300 –2 2 4 6 8 4. f(x) = g(x) h(x) e f(x) = (g ◦ h)(x) ambas no ponto x = 4 h g y x –2 2 4 6 8 –2 2 4 6 8 Nos exercic´ıos 5. a 10. calcule o limite, caso exista. Caso n˜ao exista, justifique. 5. lim x→+∞ ( xn − xn−1 ) 6. lim x→+∞ (x + 1)(x + 2) · · · (x + 10) (x2 + 1) 5 7. lim x→−∞ √ x2 − 2x + 2 x + 1 8. lim x→−∞ ( x + √ x2 + 3x + 2 ) 9. lim x→−1 ( 3 x + 1 − 5 x2 − 1 ) 10. lim x→5 (√ 25 − x2 x − 5 ) 11. Seja f definida por f(x) =    x3 + 2x2 + x x3 + 5x2 + 7x + 3 se x ̸= −3, x ̸= −1 0 se x = −3 −1/2 se x = −1 (a) A fun¸c˜ao f est´a definida em R? Justifique. (b) Dˆe os pontos onde f ´e cont´ınua. Justifique. (c) Dˆe os pontos onde f ´e descont´ınua. Justifique. (d) A fun¸c˜ao f ´e cont´ınua em R? Justifique. Nos exerc´ıcios 12. a 15. determine as equa¸c˜oes das ass´ıntotas verticais e horizontais do gr´afico da fun¸c˜ao dada. 12. f(x) = 3x x − 1 13. f(x) = 2x √ x2 + 4 14. f(x) = 2x2 + 1 2x2 − 3x 15. f(x) = x √ x2 − 4 16. A fun¸c˜ao f ´e tal que para x ̸= 2, f satisfaz 1 + 4x − x2 ≤ f(x) ≤ x2 − 4x + 9. Calcule lim x→2 f(x).
  • 2. Lista 4 de C´alculo I 2010-2 8 17. Seja f uma fun¸c˜ao limitada. Use o teorema do anulamento (´e o corol´ario do teorema do confronto) para provar que lim x→0 x2 f(x) = 0. 18. Sabendo que para x > 1, f(x) satisfaz (x − 1)2 < ( x2 − 1 ) · f(x) < (x + 1)2 , calcule lim x→+∞ f(x). Nos exercic´ıos 19. a 27. calcule o limite, caso exista. Caso n˜ao exista, justifique. 19. lim x→0 sen x3 x 20. lim x→0 tan(πx) tan x 21. lim x→0 sen 2 ( ax2 ) x4 22. lim x→0 1 − cos(ax) x2 23. lim x→0 1 − sec x x2 24. lim x→0 sen (x) sen (3x) sen (5x) tan(2x) tan(4x) tan(6x) 25. lim x→0 √ 1 + tan x − √ 1 + sen x x3 26. lim x→0 ( x cos 1 x ) 27. lim x→−2 ( x2 − 4 ) sen ( 1 x + 2 ) 28. lim x→+∞ x − cos x x 29. lim x→−∞ 1 + x sen x x 30. lim x→−∞ x2 sen x Nos exerc´ıcios 31. a 33. verifique se a fun¸c˜ao dada tem extens˜ao cont´ınua a toda reta R. 31. f(x) = sen 2 4x x 32. f(x) = −1 + sen x x − π/2 33. f(x) = sen ( x2 − 4 ) x + 2 RESPOSTAS 1. lim x→2− f(x) = +∞; lim x→2+ f(x) = −∞ 2. lim x→3− f(x) = 0; lim x→3+ f(x) = −∞ 3. lim x→2− f(x) = 0; lim x→2+ f(x) = 0 4. lim x→4− g(x) h(x) = lim x→4+ g(x) h(x) = −∞ lim x→4− (g ◦ h)(x) = lim x→4+ (g ◦ h)(x) = 5 5. , pois quando x → +∞ a fun¸c˜ao → +∞ 6. 1 7. −1 8. − 3 2 9. , pois a fun¸c˜ao → −∞ se x → −1− (ou, a fun¸c˜ao → +∞ se x → −1+ ) 10. , pois a fun¸c˜ao → −∞ se x → 5− . Obs.: ̸ ∃x; x → 5+ , pois neste caso −5 ≤ x < 5. 11. (a) Sim, pois a ´unica restri¸c˜ao da express˜ao ´e o denominador n˜ao nulo, os ´unicos pontos que anulam o denominador s˜ao x = −1 e x = −3 e nestes pontos a fun¸c˜ao foi definida por outras express˜oes, a saber f(−1) = −1/2 e f(−3) = 0. (b) Em R − {−3, −1} a fun¸c˜ao ´e cont´ınua pois ´e o quociente de fun¸c˜oes polinomiais e toda fun¸c˜ao polinomial ´e cont´ınua. Em x = −1 a fun¸c˜ao ´e cont´ınua pois lim x→−1 f(x) = − 1 2 = f(−1). (c) A fun¸c˜ao ´e descont´ınua em x = −3 pois f(x) → +∞ se x → −3− (outra justificativa seria f(x) → −∞ se x → −3+ , basta n˜ao ter um dos limites laterais). (d) N˜ao, pois n˜ao ´e cont´ınua em x = −3. 12. V: x = 1; H: y = 3 13. V: n˜ao tem; H: y = −2, y = 2 14. V: x = 0, x = 3 2 ; H: y = 1 15. V: x = −2, x = 2; H: y = −1, y = 1 16. 5 17. (i) Para g(x) = x2 e a = 0, temos lim x→a g(x) = lim x→0 x2 = 0 (ii) f ´e limitada, isto significa que ∃M; |f(x)| ≤ M. Assim, as duas hip´oteses (i) e (ii) do teorema do anula- mento se verificam. Logo vale a tese do teorema, a saber lim x→a g(x)f(x) = 0 ⇒ lim x→0 x2 f(x) = 0. 18. 1 19. 0 20. π 21. a2 22. a2 2 23. − 1 2 24. 5 16 25. 1 4 26. 0 27. 0 28. 1 29. , oscila entre −1 e 1 30. , oscila entre −∞ e +∞ 31. Sim, g(x) =    sen 2 4x x , x ̸= 0 0 , x = 0 32. Sim, g(x) =    −1 + sen x x − π/2 , x ̸= π/2 0 , x = π/2 33. Sim, g(x) =    sen ( x2 − 4 ) x + 2 , x ̸= −2 −4 , x = −2

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